Proving Leibniz theorem using induction 
Let $f$ and $g$ be $n$ times differentiable functions then prove that $$\frac {d^n}{dx^n}(fg)=\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)} $$ where $f^{(k)} $ is $k$-th  derivative with respect to $x $. 

Now I started with Mathematical Induction. I know its true for $n=1$ so skipped it.
Let it be true for $m<n $ thus $$\frac {d^m}{dx^m}(fg)=\sum_{i=0}^{m} \binom {m}{i}f^{(m)}g^{(m-i)} =s .$$ We need to prove this for $m+1$ . Note that $m+1 <n $ is also true. So we see that 
$$\frac {d^{m+1}}{dx^{m+1}}(fg)=\frac {d}{dx}(s).$$ But now problem here is that I don't know how to differentiate this whole $s$ as we cant use Leibniz rule to prove Leibniz rule.
 A: Since 
$$\frac {d^{m+1}}{dx^{m+1}}(fg)=\frac {d^{m}}{dx^{m}}\left(\frac {d}{dx}(fg)\right)=\frac {d^{m}}{dx^{m}}(f'g)+\frac {d^{m}}{dx^{m}}(fg')$$
apply Leibniz rule of order $m$ to the products $f'g$ and $fg'$.
Hence
\begin{align*}
\frac {d^{m+1}}{dx^{m+1}}(fg)&=\sum_{i=0}^{m} \binom {m}{i}f^{(i+1)}g^{(m-i)}+\sum_{i=0}^{m} \binom {m}{i}f^{(i)}g^{(m+1-i)}\\
&=f^{(m+1)}g+
\sum_{i=1}^{m} \binom {m}{i-1}f^{(i)}g^{(m+1-i)}
+\sum_{i=1}^{m} \binom {m}{i}f^{(i)}g^{(m+1-i)}+fg^{(m+1)}\\
&=f^{(m+1)}g+
\sum_{i=1}^{m} \left(\binom {m}{i-1}+\binom {m}{i}\right) f^{(i)}g^{(m+1-i)}+fg^{(m+1)}\\
&=
\sum_{i=0}^{m+1} \binom {m+1}{i} f^{(i)}g^{(m+1-i)}.
\end{align*}
where we shifted the index $i$ in the first sum.
A: I would go a different way
than Robert Z. does.
If
$\frac {d^n}{dx^n}(fg)
=\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)}
$
then
$\begin{array}\\
\frac {d^{n+1}}{dx^{n+1}}(fg)
&=\left(\frac {d^n}{dx^n}(fg)\right)'\\
&=\left(\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)}\right)'\\
&=\sum _{i=0}^n \binom{n}{ i} \left(f^{(i)}g^{(n-i)}\right)'\\
&=\sum _{i=0}^n \binom{n}{ i} \left(f^{(i+1)}g^{(n-i)}+f^{(i)}g^{(n-i+1)}\right)\\
&=\sum _{i=0}^n \binom{n}{ i} f^{(i+1)}g^{(n-i)}+\sum _{i=0}^n \binom{n}{ i}f^{(i)}g^{(n-i+1)}\\
&=\sum _{i=1}^{n+1} \binom{n}{ i-1} f^{(i)}g^{(n-i+1)}+\sum _{i=0}^n \binom{n}{ i}f^{(i)}g^{(n-i+1)}\\
&=\sum _{i=1}^{n} \binom{n}{ i-1} f^{(i)}g^{(n-i+1)}+\binom{n}{ n} f^{(n+1)}g^{(0)}+\sum _{i=1}^n \binom{n}{ i}f^{(i)}g^{(n-i+1)}+\binom{n}{ 0}f^{(0)}g^{(n+1)}\\
&=f^{(n+1)}g+fg^{(n+1)}+\sum _{i=1}^{n}  f^{(i)}g^{(n-i+1)}(\binom{n}{ i-1}+\binom{n}{ i})\\
&=f^{(n+1)}g+fg^{(n+1)}+\sum _{i=1}^{n}  f^{(i)}g^{(n-i+1)}\binom{n+1}{ i})\\
&=\sum _{i=0}^{n+1}  f^{(i)}g^{(n-i+1)}\binom{n+1}{ i})\\
\end{array}
$
Either way,
the derivative of a product
is used.
